JEE Main 2022 Question Paper 26 June Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1 
FINAL JEE–MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)               TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
 Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
 
 
SECTION-A 
1. An expression for a dimensionless quantity P is
given by 
e
kt
P log
x
æö a
=
ç÷
bb
èø
; where a and b are 
constants, x is distance ; k is Boltzmann constant 
and t is the temperature. Then the dimensions of a 
will be : 
(A) [M
0
L
–1
T
0
] (B) [ML
0
T
–2
]
(C) [MLT
–2
] (D) [ML
2
T
–2
]
Official Ans. by NTA (C)
Sol. 
e
kt
P log
x
æö a
=
ç÷
bb
èø
 
kt
1
x
=
b
 Þ 
22
kt MLT
xL
-
b==
1
E kt
2
æö
=
ç÷
èø
Q
As P is dimensionless 
Þ [a] = [b] = [MLT
–2
] 
2. A person is standing in an elevator. In which
situation, he experiences weight loss ?
(A)When the elevator moves upward with constant
acceleration 
(B) When the elevator moves downward with
constant acceleration
(C) When the elevator moves upward with uniform
velocity
(D)When the elevator moves downward with  
uniform velocity 
Official Ans. by NTA (B) 
 
Sol. 
mg 
a 
N 
mg – N = ma 
Þ N = m(g – a) 
\ Person experiences weightloss, when 
acceleration of lift is downward. 
3. An object is thrown vertically upwards. At its
maximum height, which of the following quantity
becomes zero ?
(A) Momentum (B) Potential energy
(C) Acceleration (D) Force
Official Ans. by NTA (A)
Sol. At maximum height, V = 0 
\ Momentum of object is zero. 
4. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
The ratio of the centripetal force and normal
reaction on the ball at point Q is A while angular
position of point Q is a with respect to point P.
Which of the following graphs represent the
correct relation between A and a when ball goes
from Q to R ?
O
Q
R 
P
a
(A) 
A
a
 (B) 
A
a
(C) 
A
a
(D) 
A
a
Official Ans. by NTA (C) 
 
Page 2


1 
FINAL JEE–MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)               TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
 Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
 
 
SECTION-A 
1. An expression for a dimensionless quantity P is
given by 
e
kt
P log
x
æö a
=
ç÷
bb
èø
; where a and b are 
constants, x is distance ; k is Boltzmann constant 
and t is the temperature. Then the dimensions of a 
will be : 
(A) [M
0
L
–1
T
0
] (B) [ML
0
T
–2
]
(C) [MLT
–2
] (D) [ML
2
T
–2
]
Official Ans. by NTA (C)
Sol. 
e
kt
P log
x
æö a
=
ç÷
bb
èø
 
kt
1
x
=
b
 Þ 
22
kt MLT
xL
-
b==
1
E kt
2
æö
=
ç÷
èø
Q
As P is dimensionless 
Þ [a] = [b] = [MLT
–2
] 
2. A person is standing in an elevator. In which
situation, he experiences weight loss ?
(A)When the elevator moves upward with constant
acceleration 
(B) When the elevator moves downward with
constant acceleration
(C) When the elevator moves upward with uniform
velocity
(D)When the elevator moves downward with  
uniform velocity 
Official Ans. by NTA (B) 
 
Sol. 
mg 
a 
N 
mg – N = ma 
Þ N = m(g – a) 
\ Person experiences weightloss, when 
acceleration of lift is downward. 
3. An object is thrown vertically upwards. At its
maximum height, which of the following quantity
becomes zero ?
(A) Momentum (B) Potential energy
(C) Acceleration (D) Force
Official Ans. by NTA (A)
Sol. At maximum height, V = 0 
\ Momentum of object is zero. 
4. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
The ratio of the centripetal force and normal
reaction on the ball at point Q is A while angular
position of point Q is a with respect to point P.
Which of the following graphs represent the
correct relation between A and a when ball goes
from Q to R ?
O
Q
R 
P
a
(A) 
A
a
 (B) 
A
a
(C) 
A
a
(D) 
A
a
Official Ans. by NTA (C) 
 
 
2 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. V 2gRsin =a 
 N – mg sin a = 
2
mv
2mgsin
R
=a 
 
 
mg 
N 
a 
a 
 
 
N 13
1
2mgsin 22
= +=
a
 
 
a 
A
 
 Þ A = constant 
5. A thin circular ring of mass M and radius R is 
rotating with a constant angular velocity 2 rads
–1
 in 
a horizontal plane about an axis vertical to its plane 
and passing through the center of the ring. If two 
objects each of mass m be attached gently to the 
opposite ends of a diameter of ring, the ring will 
then rotate with an angular velocity (in rads
–1
). 
 (A) 
( )
M
Mm +
  (B) 
( ) M 2m
2M
+
 
 (C) 
( )
2M
M 2m +
 (D) 
( ) 2 M 2m
M
+
 
 Official Ans. by NTA (C) 
  
Sol. Applying conservation of angular momentum  
 MR
2
w = (MR
2
 + 2mR
2
)w' 
 
2M
'
M 2m
w=
+
 
6. The variation of acceleration due to gravity (g) 
with distance (r) from the center of the earth is 
correctly represented by : (Given R = radius of 
earth) 
 (A) 
g
O
r
R 
   
 (B) 
 
g
O
r
R
 
 
 (C) 
 
g
O
r
R
   
 (D) 
 
g
O
r
R 
 
 Official Ans. by NTA (A) 
  
Page 3


1 
FINAL JEE–MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)               TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
 Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
 
 
SECTION-A 
1. An expression for a dimensionless quantity P is
given by 
e
kt
P log
x
æö a
=
ç÷
bb
èø
; where a and b are 
constants, x is distance ; k is Boltzmann constant 
and t is the temperature. Then the dimensions of a 
will be : 
(A) [M
0
L
–1
T
0
] (B) [ML
0
T
–2
]
(C) [MLT
–2
] (D) [ML
2
T
–2
]
Official Ans. by NTA (C)
Sol. 
e
kt
P log
x
æö a
=
ç÷
bb
èø
 
kt
1
x
=
b
 Þ 
22
kt MLT
xL
-
b==
1
E kt
2
æö
=
ç÷
èø
Q
As P is dimensionless 
Þ [a] = [b] = [MLT
–2
] 
2. A person is standing in an elevator. In which
situation, he experiences weight loss ?
(A)When the elevator moves upward with constant
acceleration 
(B) When the elevator moves downward with
constant acceleration
(C) When the elevator moves upward with uniform
velocity
(D)When the elevator moves downward with  
uniform velocity 
Official Ans. by NTA (B) 
 
Sol. 
mg 
a 
N 
mg – N = ma 
Þ N = m(g – a) 
\ Person experiences weightloss, when 
acceleration of lift is downward. 
3. An object is thrown vertically upwards. At its
maximum height, which of the following quantity
becomes zero ?
(A) Momentum (B) Potential energy
(C) Acceleration (D) Force
Official Ans. by NTA (A)
Sol. At maximum height, V = 0 
\ Momentum of object is zero. 
4. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
The ratio of the centripetal force and normal
reaction on the ball at point Q is A while angular
position of point Q is a with respect to point P.
Which of the following graphs represent the
correct relation between A and a when ball goes
from Q to R ?
O
Q
R 
P
a
(A) 
A
a
 (B) 
A
a
(C) 
A
a
(D) 
A
a
Official Ans. by NTA (C) 
 
 
2 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. V 2gRsin =a 
 N – mg sin a = 
2
mv
2mgsin
R
=a 
 
 
mg 
N 
a 
a 
 
 
N 13
1
2mgsin 22
= +=
a
 
 
a 
A
 
 Þ A = constant 
5. A thin circular ring of mass M and radius R is 
rotating with a constant angular velocity 2 rads
–1
 in 
a horizontal plane about an axis vertical to its plane 
and passing through the center of the ring. If two 
objects each of mass m be attached gently to the 
opposite ends of a diameter of ring, the ring will 
then rotate with an angular velocity (in rads
–1
). 
 (A) 
( )
M
Mm +
  (B) 
( ) M 2m
2M
+
 
 (C) 
( )
2M
M 2m +
 (D) 
( ) 2 M 2m
M
+
 
 Official Ans. by NTA (C) 
  
Sol. Applying conservation of angular momentum  
 MR
2
w = (MR
2
 + 2mR
2
)w' 
 
2M
'
M 2m
w=
+
 
6. The variation of acceleration due to gravity (g) 
with distance (r) from the center of the earth is 
correctly represented by : (Given R = radius of 
earth) 
 (A) 
g
O
r
R 
   
 (B) 
 
g
O
r
R
 
 
 (C) 
 
g
O
r
R
   
 (D) 
 
g
O
r
R 
 
 Official Ans. by NTA (A) 
  
 
3 
Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. 
3
2
GMr
,rR
R
g
GM
,rR
r
ì
£
ï
ï
=
í
ï
³
ï
î
g
O
r
R 
7. The efficiency of a Carnot's engine, working
between steam point and ice point, will be :
(A) 26.81% (B) 37.81%
(C) 47.81% (D) 57.81%
Official Ans. by NTA (A) 
 
Sol. 
L
n
T
1 100%
T
éù
h= -´
êú
ëû
T
L
 = 0°C = 273K, T
n
 = 373 K 
\ h = 26.809% 
8. Time period of a simple pendulum in a stationary
lift is 'T'. If the lift accelerates with 
g
6
 vertically 
upwards then the time period will be : 
(where g = acceleration due to gravity) 
(A) 
6
T
5
(B) 
5
T
6
(C) 
6
T
7
(D) 
7
T
6
Official Ans. by NTA (C) 
 
Sol. 
eff
T2
g
=p
l
a 
(a) when a = 0, T = 2
g
p
l
(b) when a =
g
6
, T'2
g
g
6
=p
+
l
\ T' = 
6
T
7
9. A thermally insulated vessel contains an ideal gas
of molecular mass M and ratio of specific heats
1.4. Vessel is moving with speed v and is suddenly
brought to rest. Assuming no heat is lost to the
surrounding and vessel temperature of the gas
increases by : (R = universal gas constant)
(A) 
2
Mv
7R
(B) 
2
Mv
5R
(C) 
2
Mv
2
7R
(D) 
2
Mv
7
5R
Official Ans. by NTA (B) 
 
Sol. 
P
v
C 2
1 1.4
CF
= += Þ F = 5 
By conservation of energy  
[]
2
F1
nR T nmv
22
D=
DT = 
22
mv Mv
FR 5R
=
Page 4


1 
FINAL JEE–MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)               TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
 Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
 
 
SECTION-A 
1. An expression for a dimensionless quantity P is
given by 
e
kt
P log
x
æö a
=
ç÷
bb
èø
; where a and b are 
constants, x is distance ; k is Boltzmann constant 
and t is the temperature. Then the dimensions of a 
will be : 
(A) [M
0
L
–1
T
0
] (B) [ML
0
T
–2
]
(C) [MLT
–2
] (D) [ML
2
T
–2
]
Official Ans. by NTA (C)
Sol. 
e
kt
P log
x
æö a
=
ç÷
bb
èø
 
kt
1
x
=
b
 Þ 
22
kt MLT
xL
-
b==
1
E kt
2
æö
=
ç÷
èø
Q
As P is dimensionless 
Þ [a] = [b] = [MLT
–2
] 
2. A person is standing in an elevator. In which
situation, he experiences weight loss ?
(A)When the elevator moves upward with constant
acceleration 
(B) When the elevator moves downward with
constant acceleration
(C) When the elevator moves upward with uniform
velocity
(D)When the elevator moves downward with  
uniform velocity 
Official Ans. by NTA (B) 
 
Sol. 
mg 
a 
N 
mg – N = ma 
Þ N = m(g – a) 
\ Person experiences weightloss, when 
acceleration of lift is downward. 
3. An object is thrown vertically upwards. At its
maximum height, which of the following quantity
becomes zero ?
(A) Momentum (B) Potential energy
(C) Acceleration (D) Force
Official Ans. by NTA (A)
Sol. At maximum height, V = 0 
\ Momentum of object is zero. 
4. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
The ratio of the centripetal force and normal
reaction on the ball at point Q is A while angular
position of point Q is a with respect to point P.
Which of the following graphs represent the
correct relation between A and a when ball goes
from Q to R ?
O
Q
R 
P
a
(A) 
A
a
 (B) 
A
a
(C) 
A
a
(D) 
A
a
Official Ans. by NTA (C) 
 
 
2 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. V 2gRsin =a 
 N – mg sin a = 
2
mv
2mgsin
R
=a 
 
 
mg 
N 
a 
a 
 
 
N 13
1
2mgsin 22
= +=
a
 
 
a 
A
 
 Þ A = constant 
5. A thin circular ring of mass M and radius R is 
rotating with a constant angular velocity 2 rads
–1
 in 
a horizontal plane about an axis vertical to its plane 
and passing through the center of the ring. If two 
objects each of mass m be attached gently to the 
opposite ends of a diameter of ring, the ring will 
then rotate with an angular velocity (in rads
–1
). 
 (A) 
( )
M
Mm +
  (B) 
( ) M 2m
2M
+
 
 (C) 
( )
2M
M 2m +
 (D) 
( ) 2 M 2m
M
+
 
 Official Ans. by NTA (C) 
  
Sol. Applying conservation of angular momentum  
 MR
2
w = (MR
2
 + 2mR
2
)w' 
 
2M
'
M 2m
w=
+
 
6. The variation of acceleration due to gravity (g) 
with distance (r) from the center of the earth is 
correctly represented by : (Given R = radius of 
earth) 
 (A) 
g
O
r
R 
   
 (B) 
 
g
O
r
R
 
 
 (C) 
 
g
O
r
R
   
 (D) 
 
g
O
r
R 
 
 Official Ans. by NTA (A) 
  
 
3 
Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. 
3
2
GMr
,rR
R
g
GM
,rR
r
ì
£
ï
ï
=
í
ï
³
ï
î
g
O
r
R 
7. The efficiency of a Carnot's engine, working
between steam point and ice point, will be :
(A) 26.81% (B) 37.81%
(C) 47.81% (D) 57.81%
Official Ans. by NTA (A) 
 
Sol. 
L
n
T
1 100%
T
éù
h= -´
êú
ëû
T
L
 = 0°C = 273K, T
n
 = 373 K 
\ h = 26.809% 
8. Time period of a simple pendulum in a stationary
lift is 'T'. If the lift accelerates with 
g
6
 vertically 
upwards then the time period will be : 
(where g = acceleration due to gravity) 
(A) 
6
T
5
(B) 
5
T
6
(C) 
6
T
7
(D) 
7
T
6
Official Ans. by NTA (C) 
 
Sol. 
eff
T2
g
=p
l
a 
(a) when a = 0, T = 2
g
p
l
(b) when a =
g
6
, T'2
g
g
6
=p
+
l
\ T' = 
6
T
7
9. A thermally insulated vessel contains an ideal gas
of molecular mass M and ratio of specific heats
1.4. Vessel is moving with speed v and is suddenly
brought to rest. Assuming no heat is lost to the
surrounding and vessel temperature of the gas
increases by : (R = universal gas constant)
(A) 
2
Mv
7R
(B) 
2
Mv
5R
(C) 
2
Mv
2
7R
(D) 
2
Mv
7
5R
Official Ans. by NTA (B) 
 
Sol. 
P
v
C 2
1 1.4
CF
= += Þ F = 5 
By conservation of energy  
[]
2
F1
nR T nmv
22
D=
DT = 
22
mv Mv
FR 5R
=
 
4 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
10. Two capacitors having capacitance C
1
 and C
2
 
respectively are connected as shown in figure. 
Initially, capacitor C
1
 is charged to a potential 
difference V volt by a battery. The battery is then 
removed and the charged capacitor C
1
 is now 
connected to uncharged capacitor C
2
 by closing the 
switch S. The amount of charge on the capacitor 
C
2
, after equilibrium is : 
 
S
C
1
 C
2
 
 (A) 
( )
12
12
CC
V
CC +
 (B) 
( )
12
12
CC
V
CC
+
 
 (C) (C
1
 + C
2
)V (D) (C
1
 – C
2
)V 
 Official Ans. by NTA (A) 
  
Sol. Charge on capacitor C
2 
 
= 
[ ]
21 2 total
total 12
C CV
CQ
C CC
´
=
+
 = 
12
12
CCV
CC +
 
11. Assertion (A) : Non-polar amterials do not have 
my permanent dipole moment. 
 Reason (R) : When an non-polar material is placed 
in a electric field. the centre of the positive charge 
distribution of it's individual atom or molecule 
coinsides with the centre of the negative charge 
distribution. 
 In the light of above statements, choose the most 
appropriate answer from the options given below. 
 (A) Both (A) and (R) are correct and (R) is the 
correct explanation of (A). 
 (B) Both (A) and (R) are correct and (R) is not the  
correct explanation of (A). 
 (C) (A) is correct but (R) is not correct. 
 (D) (A) is not correct but (R) is correct. 
 Official Ans. by NTA (C) 
  
Sol. S1 : In nonpolar molecules, centre of +ve charge  
coincides with centre of –ve charge, hence net 
dipole moment is comes to zero. 
 S2 : When non polar material is placed in external 
field, centre of charges does not coincide, hence 
give non zero moment in field 
12. The magnetic flux through a coil perpendicular to 
its plane is varying according to the relation f = 
(5t
3
 + 4t + 2t – 5) Weber. If the resistant of the coil 
is 5 ohm, then the induced current through the coil 
at t = 2 sec will be: 
 (A) 15.6 A  (B) 16.6 A 
 (C) 17.6 A  (D) 18.6 A 
 Official Ans. by NTA (A) 
  
Sol. f = 5t
3
 + 4t
2
 + 2t – 5 
 
2
d
e 15t 8t 2
dt
f
= = ++ 
 At t = 2, |e| = 15 × 2
2
 + 8 ×  2 + 2 
 Þ e = 78V Þ I = 
e 78
15.60
R5
== 
13. An aluminium wire is stretched to make its length, 
04% larger. Then percentage change in resistance 
is:  
 (A) 0.4 %  (B) 0.2 % 
 (C) 0.8 %  (D) 0.6 % 
 Official Ans. by NTA (C) 
  
Sol. 
r
=
l
R
A
 
 
RA
RA
D DD
=-
l
l
 
 lA = k 
 
A
0
A
DD
+=
l
l
 
 
R2
R
DD
=
l
l
 
 
R
2 0.4 0.8%
R
D
=´= 
Page 5


1 
FINAL JEE–MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)               TIME : 9 : 00 AM  to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
 Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
 
 
SECTION-A 
1. An expression for a dimensionless quantity P is
given by 
e
kt
P log
x
æö a
=
ç÷
bb
èø
; where a and b are 
constants, x is distance ; k is Boltzmann constant 
and t is the temperature. Then the dimensions of a 
will be : 
(A) [M
0
L
–1
T
0
] (B) [ML
0
T
–2
]
(C) [MLT
–2
] (D) [ML
2
T
–2
]
Official Ans. by NTA (C)
Sol. 
e
kt
P log
x
æö a
=
ç÷
bb
èø
 
kt
1
x
=
b
 Þ 
22
kt MLT
xL
-
b==
1
E kt
2
æö
=
ç÷
èø
Q
As P is dimensionless 
Þ [a] = [b] = [MLT
–2
] 
2. A person is standing in an elevator. In which
situation, he experiences weight loss ?
(A)When the elevator moves upward with constant
acceleration 
(B) When the elevator moves downward with
constant acceleration
(C) When the elevator moves upward with uniform
velocity
(D)When the elevator moves downward with  
uniform velocity 
Official Ans. by NTA (B) 
 
Sol. 
mg 
a 
N 
mg – N = ma 
Þ N = m(g – a) 
\ Person experiences weightloss, when 
acceleration of lift is downward. 
3. An object is thrown vertically upwards. At its
maximum height, which of the following quantity
becomes zero ?
(A) Momentum (B) Potential energy
(C) Acceleration (D) Force
Official Ans. by NTA (A)
Sol. At maximum height, V = 0 
\ Momentum of object is zero. 
4. A ball is released from rest from point P of a
smooth semi-spherical vessel as shown in figure.
The ratio of the centripetal force and normal
reaction on the ball at point Q is A while angular
position of point Q is a with respect to point P.
Which of the following graphs represent the
correct relation between A and a when ball goes
from Q to R ?
O
Q
R 
P
a
(A) 
A
a
 (B) 
A
a
(C) 
A
a
(D) 
A
a
Official Ans. by NTA (C) 
 
 
2 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. V 2gRsin =a 
 N – mg sin a = 
2
mv
2mgsin
R
=a 
 
 
mg 
N 
a 
a 
 
 
N 13
1
2mgsin 22
= +=
a
 
 
a 
A
 
 Þ A = constant 
5. A thin circular ring of mass M and radius R is 
rotating with a constant angular velocity 2 rads
–1
 in 
a horizontal plane about an axis vertical to its plane 
and passing through the center of the ring. If two 
objects each of mass m be attached gently to the 
opposite ends of a diameter of ring, the ring will 
then rotate with an angular velocity (in rads
–1
). 
 (A) 
( )
M
Mm +
  (B) 
( ) M 2m
2M
+
 
 (C) 
( )
2M
M 2m +
 (D) 
( ) 2 M 2m
M
+
 
 Official Ans. by NTA (C) 
  
Sol. Applying conservation of angular momentum  
 MR
2
w = (MR
2
 + 2mR
2
)w' 
 
2M
'
M 2m
w=
+
 
6. The variation of acceleration due to gravity (g) 
with distance (r) from the center of the earth is 
correctly represented by : (Given R = radius of 
earth) 
 (A) 
g
O
r
R 
   
 (B) 
 
g
O
r
R
 
 
 (C) 
 
g
O
r
R
   
 (D) 
 
g
O
r
R 
 
 Official Ans. by NTA (A) 
  
 
3 
Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
Sol. 
3
2
GMr
,rR
R
g
GM
,rR
r
ì
£
ï
ï
=
í
ï
³
ï
î
g
O
r
R 
7. The efficiency of a Carnot's engine, working
between steam point and ice point, will be :
(A) 26.81% (B) 37.81%
(C) 47.81% (D) 57.81%
Official Ans. by NTA (A) 
 
Sol. 
L
n
T
1 100%
T
éù
h= -´
êú
ëû
T
L
 = 0°C = 273K, T
n
 = 373 K 
\ h = 26.809% 
8. Time period of a simple pendulum in a stationary
lift is 'T'. If the lift accelerates with 
g
6
 vertically 
upwards then the time period will be : 
(where g = acceleration due to gravity) 
(A) 
6
T
5
(B) 
5
T
6
(C) 
6
T
7
(D) 
7
T
6
Official Ans. by NTA (C) 
 
Sol. 
eff
T2
g
=p
l
a 
(a) when a = 0, T = 2
g
p
l
(b) when a =
g
6
, T'2
g
g
6
=p
+
l
\ T' = 
6
T
7
9. A thermally insulated vessel contains an ideal gas
of molecular mass M and ratio of specific heats
1.4. Vessel is moving with speed v and is suddenly
brought to rest. Assuming no heat is lost to the
surrounding and vessel temperature of the gas
increases by : (R = universal gas constant)
(A) 
2
Mv
7R
(B) 
2
Mv
5R
(C) 
2
Mv
2
7R
(D) 
2
Mv
7
5R
Official Ans. by NTA (B) 
 
Sol. 
P
v
C 2
1 1.4
CF
= += Þ F = 5 
By conservation of energy  
[]
2
F1
nR T nmv
22
D=
DT = 
22
mv Mv
FR 5R
=
 
4 
          Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
10. Two capacitors having capacitance C
1
 and C
2
 
respectively are connected as shown in figure. 
Initially, capacitor C
1
 is charged to a potential 
difference V volt by a battery. The battery is then 
removed and the charged capacitor C
1
 is now 
connected to uncharged capacitor C
2
 by closing the 
switch S. The amount of charge on the capacitor 
C
2
, after equilibrium is : 
 
S
C
1
 C
2
 
 (A) 
( )
12
12
CC
V
CC +
 (B) 
( )
12
12
CC
V
CC
+
 
 (C) (C
1
 + C
2
)V (D) (C
1
 – C
2
)V 
 Official Ans. by NTA (A) 
  
Sol. Charge on capacitor C
2 
 
= 
[ ]
21 2 total
total 12
C CV
CQ
C CC
´
=
+
 = 
12
12
CCV
CC +
 
11. Assertion (A) : Non-polar amterials do not have 
my permanent dipole moment. 
 Reason (R) : When an non-polar material is placed 
in a electric field. the centre of the positive charge 
distribution of it's individual atom or molecule 
coinsides with the centre of the negative charge 
distribution. 
 In the light of above statements, choose the most 
appropriate answer from the options given below. 
 (A) Both (A) and (R) are correct and (R) is the 
correct explanation of (A). 
 (B) Both (A) and (R) are correct and (R) is not the  
correct explanation of (A). 
 (C) (A) is correct but (R) is not correct. 
 (D) (A) is not correct but (R) is correct. 
 Official Ans. by NTA (C) 
  
Sol. S1 : In nonpolar molecules, centre of +ve charge  
coincides with centre of –ve charge, hence net 
dipole moment is comes to zero. 
 S2 : When non polar material is placed in external 
field, centre of charges does not coincide, hence 
give non zero moment in field 
12. The magnetic flux through a coil perpendicular to 
its plane is varying according to the relation f = 
(5t
3
 + 4t + 2t – 5) Weber. If the resistant of the coil 
is 5 ohm, then the induced current through the coil 
at t = 2 sec will be: 
 (A) 15.6 A  (B) 16.6 A 
 (C) 17.6 A  (D) 18.6 A 
 Official Ans. by NTA (A) 
  
Sol. f = 5t
3
 + 4t
2
 + 2t – 5 
 
2
d
e 15t 8t 2
dt
f
= = ++ 
 At t = 2, |e| = 15 × 2
2
 + 8 ×  2 + 2 
 Þ e = 78V Þ I = 
e 78
15.60
R5
== 
13. An aluminium wire is stretched to make its length, 
04% larger. Then percentage change in resistance 
is:  
 (A) 0.4 %  (B) 0.2 % 
 (C) 0.8 %  (D) 0.6 % 
 Official Ans. by NTA (C) 
  
Sol. 
r
=
l
R
A
 
 
RA
RA
D DD
=-
l
l
 
 lA = k 
 
A
0
A
DD
+=
l
l
 
 
R2
R
DD
=
l
l
 
 
R
2 0.4 0.8%
R
D
=´= 
 
 
5 
Final JEE-Main Exam June, 2022/26-06-2022/Morning Session 
14. A proton and an alpha particle of the same enter in 
a uniform magnetic field which is acting 
perpendicular to their direction of motion. The 
ratio of the circular paths described by the alpha 
particle and proton is:  
 (A) 1 : 4  (B) 4 : 1 
 (C) 2 : 1  (D) 1 : 2 
 Official Ans. by NTA (C) 
  
Sol. 
P
PP
RM q
R Mq
aa
a
=´ 
 
P
R 41
2
R 12
a
= ´= 
15. If electric field intensity of a uniform plane electro 
magnetic wave is given as  
 E = –301.6 sin(kz – wt)
x
ˆ a + 452.4 sin(kz – wt)
y
V
ˆ
a
m
 
 Then, magnetic intensity H of this wave in Am
–1
 
will be:' 
 [Given: Speed of light in vacuum c = 3 × 10
8
 ms
–1
, 
permeability of vacuum µ
0
 = 4p × 10
–7
 NA
–2
] 
 (A) 
yx
ˆˆ
0.8sin(kz t)a 0.8sin(kz t)a + -w + -w 
 (B) 
66
yx
ˆˆ 1.0 10 sin(kz t)a 1.5 10 (kz t)a
--
+ ´ -w + ´ -w 
 (C) 
yx
ˆˆ
0.8sin(kz t)a 1.2sin(kz t)a - -w - -w 
 (D)
66
yx
ˆˆ 1.0 10 sin(kz t)a 1.5 10 sin(kz t)a
--
- ´ -w - ´ -w
Official Ans. by NTA (C) 
  
Sol. ( )( ) ( )
xy
ˆˆ E 301.6sin kz t a 452.4sin kz t a = -w - + -w
r
( )( )
y
301.6
ˆ B sinkz ta
C
= -w-
r
 
  + ( )()
x
452.4
ˆ sinkz ta
C
-w- 
 ( )()
y
0
B 301.6
ˆ H sinkz ta
C
= = -w-
mm
r
r
 
  ( )()
x
452.4
ˆ sinkz ta
C
+ -w-
m
 
 ( ) ( )
yx
ˆˆ H 0.8sin kz t a 1.2sin kz t a = - -w - -w
r
 
 For direction 
 EB ´
rr
 is direction of C
r
 
 For first part 
ˆ ˆˆ
E i,B? =-= 
 
ˆ ˆˆ
E Bk ´= Þ 
ˆ ˆ
Bj =- 
 Similarly for second 
 
ˆ ˆˆ
E j,B? == 
 
ˆ ˆˆ
E Bk ´= Þ 
ˆ ˆ
Bi =- 
16. In free space, an electromagnetic wave of 3 GHz of 
3 GHz frequency strikes over the edge of an object 
of size 
100
l
, where l is the wavelength of the 
wave in free space. The phenomenon, which 
happens there will be: 
 (A) Reflection (B) Refraction 
 (C) Diffraction (D)  Scattering 
 Official Ans. by NTA (D) 
  
Sol. 
a1
100
=
l
 
 For reflection  size of obstacle must be much larger 
than wavelength, for diffraction size should be 
order of wavelength. 
 Since the object is of size 
100
l
, much smaller than 
wavelength, so scattering will occur.